Algebraically find the equation of a parabola with x-intercepts at (-6+3sqrt2, 0), (-6-3sqrt2, 0) and a range of y is less than or equal to 15.

(x+6)^2=(3sqrt(2))^2=18

x^2+12x+36-18=0; x^2+12x+18=0

If ay=x^2+12x+18 then the given roots apply when y=0.

If y<15 for all x, then ay+18=(x+6)^2 and the vertex is at (-6,-18/a) representing the maximum or minimum. But 15 represents a maximum, since the range cannot exceed 15, therefore -18/a=15 and a=-18/15=-6/5. The equation of the parabola is:

6y=-5(x^2+12x+18) or y=-(5/6)x^2-10x-15.

Check: put x=-6: y=-30+60-15=15. Put (for example) x=0, y=-15; x=-5: y=-125/6+50-15=14.17<15; x=-7: y=14.17<15.